Math Problem Statement

Find the value of 7/(7×10) + 7/(10×13) + … + 7/(49×52) + 7/(52×55).

Solution

The given problem is to find the value of the series:

77×10+710×13++749×52+752×55\frac{7}{7 \times 10} + \frac{7}{10 \times 13} + \cdots + \frac{7}{49 \times 52} + \frac{7}{52 \times 55}

To evaluate this, let's examine each term in the series. Notice that each term has the form:

7n(n+3)\frac{7}{n(n+3)}

where nn follows a pattern of values: 7,10,13,,527, 10, 13, \ldots, 52.

Step 1: Simplify Each Term

We can simplify each term 7n(n+3)\frac{7}{n(n+3)} by using partial fraction decomposition:

7n(n+3)=An+Bn+3\frac{7}{n(n+3)} = \frac{A}{n} + \frac{B}{n+3}

Multiplying through by n(n+3)n(n+3) gives:

7=A(n+3)+Bn7 = A(n+3) + Bn

Expanding and solving for AA and BB:

7=An+3A+Bn7 = An + 3A + Bn

Equating terms, we get:

  1. A+B=0A + B = 0
  2. 3A=73A = 7

From 3A=73A = 7, we find A=73A = \frac{7}{3}. Substituting into A+B=0A + B = 0, we get B=73B = -\frac{7}{3}.

Thus:

7n(n+3)=73(1n1n+3)\frac{7}{n(n+3)} = \frac{7}{3} \left( \frac{1}{n} - \frac{1}{n+3} \right)

Step 2: Rewrite the Series

Using this decomposition, the series becomes:

73(17110+110113++149152+152155)\frac{7}{3} \left( \frac{1}{7} - \frac{1}{10} + \frac{1}{10} - \frac{1}{13} + \cdots + \frac{1}{49} - \frac{1}{52} + \frac{1}{52} - \frac{1}{55} \right)

Step 3: Observe the Telescoping Nature

Notice that the terms cancel out in a telescoping manner, leaving only the first and the last terms:

73(17155)\frac{7}{3} \left( \frac{1}{7} - \frac{1}{55} \right)

Step 4: Simplify the Remaining Expression

Calculating the remaining expression:

73(17155)=73(5577×55)=7348385\frac{7}{3} \left( \frac{1}{7} - \frac{1}{55} \right) = \frac{7}{3} \left( \frac{55 - 7}{7 \times 55} \right) = \frac{7}{3} \cdot \frac{48}{385}

Simplifying further:

=7×483×385=3361155=112385= \frac{7 \times 48}{3 \times 385} = \frac{336}{1155} = \frac{112}{385}

Final Answer

The value of the series is:

112385\frac{112}{385}

Would you like a further breakdown of these steps?


Here are some related questions for further exploration:

  1. How would you approach a similar series with different constants in the denominator?
  2. Can you generalize the formula for the sum of a series in the form kn(n+m)\frac{k}{n(n+m)}?
  3. How would the answer change if the series extended to infinity?
  4. What would happen if the numerator had a different value than 7?
  5. Can you derive a telescoping series for other patterns of terms?

Tip: Telescoping series often simplify complex sums by allowing terms to cancel out, leaving only a few terms for calculation.

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Math Problem Analysis

Mathematical Concepts

Series Summation
Telescoping Series
Partial Fraction Decomposition

Formulas

Partial fraction decomposition: 7/(n(n+3)) = 7/3 * (1/n - 1/(n+3))

Theorems

Telescoping Series

Suitable Grade Level

Grades 11-12